Primes on Interval 【打表+二分】

A - Primes on Interval

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

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Description

You've decided to carry out a survey in the theory of prime numbers. Let us remind you that a prime number is a positive integer that has exactly two distinct positive integer divisors.

Consider positive integers a, a + 1, ..., b (a ≤ b). You want to find the minimum integer l (1 ≤ l ≤ b - a + 1) such that for any integerx (a ≤ x ≤ b - l + 1) among l integers x, x + 1, ..., x + l - 1 there are at least k prime numbers.

Find and print the required minimum l. If no value l meets the described limitations, print -1.

Input

A single line contains three space-separated integers a, b, k (1 ≤ a, b, k ≤ 10⁶; a ≤ b).

Output

In a single line print a single integer — the required minimum l. If there's no solution, print -1.

Sample Input

Input

2 4 2

Output

3

Input

6 13 1

Output

4

Input

1 4 3

Output

-1

代码：

#include<stdio.h>#include<string.h>int a,b,k;int dp[1000000+10],c[1000000+10];void dabiao(){memset(c,0,sizeof(c));c[0]=1;c[1]=1;for(int i=2;i<=1000000;i++){if(!c[i]){for(int j=i+i;j<=1000000;j+=i){c[j]=1;} }}dp[0]=0;for(int i=1;i<=1000000;i++){dp[i]=dp[i-1];if(!c[i])dp[i]++;} }int judge(int l){for(int i=a;i<=b-l+1;i++){if(dp[i+l-1]-dp[i-1]<k)return 0;}return 1;}int main(){dabiao();scanf("%d%d%d",&a,&b,&k);int l=1,r=b-a+1,mid,flag=0;while(l<=r){mid=(l+r)/2;if(judge(mid)){flag=1;r=mid-1;//ans=mid;}else{l=mid+1;}}if(flag)printf("%d\n",l);elseprintf("-1\n");return 0;}